Interaction of Gamma-Rays - General Considerations
uncharged
transfer of energy
creation of fast electrons
photoelectric
Compton
pair production
photodisintegration
Interaction of Gamma-Rays - Types
Photoelectric Effect
Einstein, 1905, as part of his Nobel prize winning paper on the photon theory of light - a prediction which was later verified experimentally in detail
photon absorbed by atom, goes into excited state and ejects an electron with excess kinetic energy:
smax eVK
electron ejected of energy binding
photon
hEelectron k
Photoelectric Effect
Photoelectric Effect
- h = eV
C 10 1.6 = e
potentialor voltage stopping = V
s
19
s
but can also be expressed as a work function, a constant term "eWo" which varies from material to material eVS = h - eWo
where: h (the slope) remains constant for all material being equal to Planck's constant; 6.6 10-34 J-s
there exists a threshold frequency:
hVth = eWo
below this threshold photons will not have sufficient energy to release even the least tightly bound electrons
cross-section drops abruptly at the K-edge because below there is insufficient energy to overcome the binding and the K-shell electrons no longer participate
the L-edge is really 3 energies due to fine level splitting
Photoelectric Effect
K-shell binding energies vary from 13.6 eV (H); 7.11 keV (Fe); 88 keV (Pb); 116 keV (U); if hν < Ek, only L and higher shell electrons can take part
photoelectric effect favored by low-energy photons and high Z absorbers; cross-section varies as:
)(hZ
3
4
PE
strong Z-dependence makes Pb a good x-ray absorber, usually followed by Cu and Al in a layered shield
Photoelectric Effect
as vacancy left by the photoelectron is filled by an electron from an outer shell, either fluorescence x-rays or Auger electrons may be emitted
the probability of x-ray emission is given by the "fluorescent yield"
90Zfor 0.965
8Zfor 0.005
created vacancy
emitted photonsyield tfluorescen
Photoelectric Effect
wave interpretation predicts that when electromagnetic radiation is scattered from a charged particle, the scattered radiation will have the same frequency as the incident radiation in all directions
the scattering of electromagnetic radiation from a charged particle is viewed as a perfectly elastic billiard ball
Compton Scattering
4 unknowns: E1, , K,
3 equations: momentum conservation (2), energy conservation
θ
E1 P1
K, P e-
e-
Eo Po
before after
γ
Compton Scattering
K = (m - mo)c2 difference between the total energy E of the moving particle and the rest energy Eo (at rest)
2o
2o
2o
)c m - m ( = K
cm - mc = E - E = K
must treat electron relativistically
42o
2222o
2 c m + c p = ) cm + K ( = E
Compton Scattering
while for photons
cE = p
cE = p 1
1o
o
cm+ K + E = cm + E 2o1
2oo
energy
K =(p0 – p1)c
Compton Scattering
momentum:
2
10
2
1
2
0 pcospp2pp
p0 – p1 cos = p cos
square and add
p1 sin = p sin
Compton Scattering
substitute these expressions for K1, p2 into relativistic electron energy expression to get (after manipulation):
where Δ is the shift that the scattered photon undergoes
the wavelength is usually measured in multiples of "Compton units", the ratio of:
Compton the to Ehc wavelength the
)cos1(cm
h
0
01
Compton Scattering
wavelength
1 =
)MeV ( E
.511 =
E
cm =
cm
hcE
hc
=
m 10 2.43 = cm
hc =
2e
2e
o
122
eC
) cos - 1 ( + 1
) cos - 1 ( E = E - E = K o1o
the difference in energy Eo - E1 = K is the kinetic energy of the electron
Compton Scattering
the min. electron energy corresponds to min. scattered photon energy (θ = 180º) so that
1« for E + 1
E = 2 + 1
2 E = K o
21o
omax
this energy Kmax is called the Compton edge
another form for this equation which uses photon energies instead of wavelengths is:
Compton Scattering
) cos -1 ( ) cm /E ( + 1
E = E :or
cmE =
) cos - 1 ( + 1
1 =
E
E
2o
2o
o
o
1
at high incident energies Eo the back scattered photon approaches a constant energy
1cm
Efor
cos1cm
)cos1(cm
E1
E)(E
2
0
0
2
0
2
o
0
01
Compton Scattering
0.511 MeV θ = 90º
0.255 MeV θ = 180º
in this limit we find that 0 « ≈ , so that the energy:
0
1
1 oft independen is hchc
E
Compton Scattering
In a Compton experiment an electron attains kinetic energy of 0.100 MeV when an x-ray of energy 0.500 MeV strikes it. Determine the wavelength of the scattered photon if the electron is initially at rest
A 13 = MeV0.400
A MeV10 12.4 =
E
hc =
MeV0.400 = E
MeV0.100 + E = MeV0.500
) (k E = E
E = E
3-
e
finalinitial
210.
Compton Scattering Problem
in the process of pair production the energy carried by a photon is completely converted into matter, resulting in the creation of an electron-positron pair
σpp ~ Z2
since the charge of the system was initially zero, 2 oppositely charged particles must be produced in order to conserve charge
Pair Production
in order to produce a pair, the incident photon must have an energy of the pair; any excess energy of the photon appears as kinetic energy of the particles
E = h
heavy nucleus Mo
before pair production:
Pair Production
after pair production:
E+ = m+c2
E- = m-c2
E = Moc2 + k
p+
p-
p
Pair Production
pair production cannot occur in empty space
the nucleus carries away an appreciable amount of the incident photon's momentum, but because of its large mass, its recoil kinetic energy, k ≈ p2/2mo, is usually negligible compared to kinetic energies of the electron-positron pair
thus, energy (but not momentum) conservation may be applied with the heavy nucleus ignored, yielding:
h = m+c2 + m-c2 = k+ + k- + 2moc2
since the positron and the electron have the same rest mass;
mo = 9.11 x 10-31 kg
Pair Production
the inverse of pair production can also occur
in pair annihilation a positron-electron pair is annihilated, resulting in the creation of 2 (or more) photons as shown
at least 2 photons must be produced in order to conserve energy and momentum
Annihilation
in contrast to pair production, pair annihilation can take place in empty space and both energy and momentum principles are applicable, so that:
21finalinitial
212
ofinalinitial
k 2
h + k
2
h = vm + vm or P = P
hv + hv = k + k + c2mor E = E
where k l is the propagation vector:
2 (k) = 2/
Annihilation
before pair annihilation
E+ = m+c2 = moc2 + k+
E- = m-c2 = moc2 + k-
Annihilation
after pair annihilation
h1
h2
Annihilation
problem:
how many positrons can a 200 MeV photon produce?
the energy needed to create an electron-positron pair at rest is twice the rest energy of an electron, or 1.022 MeV; therefore maximum number of positrons equals:
positrons 195 pair
positron1
MeV 1.022pair 1
MeV) 200(
Annihilation
absorber nucleus captures a -ray and in most instances emits a neutron:
9Be( ,n) 8Be
important for high energy photons from electron accelerators
cross-sections are « total cross-sections
Photodisintegration
total attenuation coefficient
ppcsPEt + + =
in computing shielding design the above equation is used
this is the fraction of the energy in a beam that is removed per unit distance of absorber
Combined Effects
the fraction of the beam's energy that is deposited in the absorber considers only the energy transferred by the photoelectron, Compton electron, and the electron pair
energy carried away by the scattered photon by Compton and by annihilation is not included
1.021.02 - hf
+ + = ppcePEe
Combined Effects
Linear Attenuation and Absorption Coefficients for Photons in Water
Exponential Absorption
due to the different interaction of -rays with matter, the attenuation is different than with α or particles
intensity of a beam of photons will be reduced as it passes through material because they will be removed or scattered by some combination of photoelectric effect, Compton scattering and pair production
reduction obeys the exponential attenuation law:
I = Ioe-t
where:I0 = -ray intensity at zero absorber
thicknesst = absorber thicknessI = -ray intensity transmitted = attenuation coefficient
if the absorber thickness is measure in cm, then is called linear attenuation coefficient (l) having dimensions "per cm"
Exponential Absorption
if the absorber thickness "t" is measured in g/cm2, then (m) is called mass attenuation coefficient (m) having dimensions cm2/g
32m
1l /cmg /gcm = cm
where is the density of the absorber
Exponential Absorption
through passes 70.5%
0.705 =
e = e = I
I
mm) (5.0 ) (0.07mm-x-
o
1
Exponential Absorption
what percent of an incident x-ray passes through a 5 mm material whose linear absorption is 0.07 mm-1?
a monochromatic beam of photons is incident on an absorbing material
if the intensity is reduced by a factor of 2 by 8 mm of material, what is the absorption coefficient?
1
mm8o
o
mm 0.0866 = for solve
e I = 2I
Exponential Absorption
Half-Value Thickness (HVT)
thickness of absorber which reduces the intensity of a photon beam to 1/2 its incident value
find HVT of aluminum if = 0.070 mm-1
mm 9.9 =x
e = 2
1
x ) mm (0.07- 1
Exponential Absorption
Atomic Attenuation Coefficient a
fraction of an incident -ray beam that is attenuated by a single atom, or the probability that an absorber atom will interact with one of the photons
where a is referred to as a cross-section and has the units barns
section-cross cmacroscopi =
) ( section-cross cmicroscopi =
cm 10 = barn 1
l
a
224
Exponential Absorption
Linear Attenuation Coefficients
what is the thickness of Al and Pb to transmit 10% of a 0.1 MeV -ray?
) less times 137 ( Pbfor cm 0.0385 =t and
) Al( cm 5.3 =t t 0.435 = 10 ln
e 10
1 = e =
I
I
Pbfor 59.7 =
for Al cm 0.435 =
cmt ) cm -(4.35t-
o
1
11
1
(0.435 cm-1)
Exponential Absorption
if we have a 1.0 MeV - ray:
) less times4.67 ( ) (Pb cm 2.97 =t
) (Al cm 13.86 =t
compute the density thickness at 0.1 MeV
23d
23d
g/cm 0.435 = cm 0.0385 g/cm 11.34 = Pb t
g/cm 14.3 = cm 5.3 g/cm 2.7 = Alt
Exponential Absorption
at 1.0 MeV
2d
2d
g/cm 33.6 = Pb t
g/cm 37.4 = Alt
this shows that Pb is only slightly better on a mass basis than Al
however for low energy photons Pb is much better
in general, for energies between 0.8 5 MeV almost all materials, on a mass basis, have approximately the same -ray attenuating properties
Exponential Absorption
1-MeV photons are normally incident on a 1-cm lead slab
the mass attenuation coefficient of lead (density = 11.35 g/cm3) is 0.0708 cm2/g and the atomic weight is 207.2
Photon Interactions - Problem
calculate the linear attenuation coefficient
what fraction of 1-MeV photons interact in a 1-cm lead slab?
what thickness of lead is required for half the incident photons to interact?
calculate the mean free path
Photon Interactions - Problem
Solution
a. the linear attenuation coefficient is obtained by multiplying the mass attenuation coefficient by the density
= (0.0708 cm2 g-1) (11.35 g cm-3 = 0.804 cm-1.
b. if Io photons are incident on the lead slab and I photons penetrate it without interacting, then the fraction not interacting is given by:
I/Io = exp[- x]
= exp[(-0.804 cm-1)(1 cm)] = 0.448
Photon Interactions - Problem
the fraction of photons interacting is then: 1 - 0.448 = 0.552
c. substituting I/Io = 1/2 in eq. (1) above and rearranging yields:
x = ln2/μ = 0.693/0.804 = 0.862 cm
Photon Interactions - Problem
d. the mean free path (MFP) is the average distance that an incident photon travels before interaction and is the reciprocal of the attenuation coefficient:
MFP = 1/μ = 1/0.804 = 1.24 cm
this is also numerically equal to the "relaxation length“
the relaxation length is the shield thickness needed to attenuate a narrow beam of monoenergetic photons to 1/e (= 0.368) of its original intensity and can be derived by substituting I/Io = 1/e
Photon Interactions - Problem